# Complementary Angles

## Complementary Angles

Theorem: $${{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2},\ for\ \ all\ \ x\in \left[ -1,1 \right]$$,

Proof: Let $${{\sin }^{-1}}x=\theta$$. . .(1)

Where $$x\in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]$$,

$$\Rightarrow -\frac{\pi }{2}\le -\theta \le \frac{\pi }{2}$$,

$$\Rightarrow -\frac{\pi }{2}+\frac{\pi }{2}\le -\theta +\frac{\pi }{2}\le \frac{\pi }{2}+\frac{\pi }{2}$$,

$$\Rightarrow 0\le \frac{\pi }{2}-\theta \le \pi$$,

$$\Rightarrow\frac{\pi }{2}-\theta \in \left[ 0,\pi \right]$$,

Now sin⁻¹x = θ

x = sin θ

x = cos (π/2 – θ)

cos⁻¹x = π/2 – θ

cos⁻¹x + θ = π/2. . . (2)

from equation 1 and 2

sin⁻¹x = θ

cos⁻¹x + θ = π/2

cos⁻¹x + sin⁻¹x = π/2

Note:

• Tan⁻¹x + cot⁻¹x = π/2, for all x ϵ R.
• sec⁻¹x + cosec⁻¹x = π/2, for all x ϵ (-∞, -1] $$\cup$$ [1, ∞).

Example: If sin⁻¹x = π/5, for some x ϵ (-1, 1), then find the value of cos⁻¹x.

Solution:

sin⁻¹x = π/5

cos⁻¹x + sin⁻¹x = π/2

cos⁻¹x + π/5= π/2

cos⁻¹x = π/2 – π/5

cos⁻¹x = 3 π/10